Question

The sum of the coefficients of \( x^{2/3} \) and \( x^{-2/5} \) in the binomial expansion of $$ \left( x^{2/3} + \frac{1}{2} x^{-2/5} \right)^9 $$ is: 

• A. \( \frac{21}{4} \)
• B. \( \frac{69}{16} \)
• C. \( \frac{63}{16} \)
• D. \( \frac{19}{4} \)

Answer 1

The Correct Option is A

To solve this problem, we need to find the sum of the coefficients of \( x^{2/3} \) and \( x^{-2/5} \) in the binomial expansion of \( \left( x^{2/3} + \frac{1}{2} x^{-2/5} \right)^9 \).

First, let's consider the binomial expansion of \( \left( a + b \right)^n \), where each term in the expansion is given by:

\(T_{k+1} = \binom{n}{k} a^{n-k} b^k\)

For this specific binomial expansion:

• \(a = x^{2/3}\)
• \(b = \frac{1}{2} x^{-2/5}\)
• \(n = 9\)

The general term in the expansion is:

\(T_{k+1} = \binom{9}{k} \left(x^{2/3}\right)^{9-k} \left(\frac{1}{2} x^{-2/5}\right)^k\)

This simplifies to:

\(T_{k+1} = \binom{9}{k} \left(x^{\frac{2}{3} \cdot (9-k)}\right) \left(\frac{1}{2}\right)^k \left(x^{-\frac{2}{5} k}\right)\)

The power of \( x \) in \( T_{k+1} \) is:

\(\frac{2}{3}(9-k) - \frac{2}{5}k\)

Simplifying,\(\frac{2}{3}(9-k) - \frac{2}{5}k = 6 - \frac{2}{3}k - \frac{2}{5}k\)

We need this power to be \( \frac{2}{3} \) and \( -\frac{2}{5} \) respectively.

Finding the coefficient for \( x^{2/3} \):

Set: \(6 - \frac{2}{3}k - \frac{2}{5}k = \frac{2}{3}\)

Solving for \( k \):

\(6 - \frac{2}{3}k - \frac{2}{5}k = \frac{2}{3}\)

\(\Rightarrow 6 - \frac{2}{3}k - \frac{2}{5}k = \frac{2}{3}\)

\(\Rightarrow 6 - \frac{10k + 6k}{15} = \frac{2}{3}\)

\(\Rightarrow 6 - \frac{16}{15}k = \frac{2}{3}\)

\(\Rightarrow \frac{16}{15}k = 6 - \frac{2}{3}\)

\(\Rightarrow \frac{16}{15}k = \frac{16}{3} \Rightarrow k = \frac{16}{3} \times \frac{15}{16} = 5\)

For \( k = 5 \), the term is:

\(T_6 = \binom{9}{5} \left(x^{2/3}\right)^4 \left(\frac{1}{2} x^{-2/5}\right)^5\)

Calculate:

\(T_6 = \binom{9}{5} \cdot x^{8/3} \cdot \frac{1}{32}x^{-2}\Rightarrow\frac{126}{32}x^{2/3}= \frac{63}{16}x^{2/3}\)

Finding the coefficient for \( x^{-2/5} \):

Set: \(6 - \frac{2}{3}k - \frac{2}{5}k = -\frac{2}{5}\)

Solving for \( k \):

\(6 - \frac{2}{3}k - \frac{2}{5}k = -\frac{2}{5}\)

\(\Rightarrow 6 - \frac{10k + 6k}{15} = -\frac{2}{5}\)

\(\Rightarrow 6 - \frac{16}{15}k = -\frac{2}{5}\)

\(\Rightarrow \frac{16}{15}k = 6 + \frac{2}{5} = \frac{32}{5}\)

\(\Rightarrow k = \frac{32}{5} \times \frac{15}{16} = 6\)

For \( k = 6 \), the term is:

\(T_7 = \binom{9}{6} \left(x^{2/3}\right)^3 \left(\frac{1}{2} x^{-2/5}\right)^6\)

Calculate:

\(T_7 = \binom{9}{6} \cdot x^{2} \cdot \frac{1}{64}x^{-12/5} \Rightarrow \frac{84}{64} x^{-2/5}= \frac{21}{16} x^{-2/5}\)

Sum of the coefficients:

Sum = Coefficient of \(x^{2/3}\) + Coefficient of \(x^{-2/5} = \frac{63}{16} + \frac{21}{16}\)

\(\Rightarrow \frac{63 + 21}{16} = \frac{84}{16} = \frac{21}{4}\)

Thus, the correct answer is \( \frac{21}{4} \).

Answer 2

Step 1. General Term of the Expansion:
The general term in the binomial expansion of \( \left( x^{2/3} + \frac{1}{2}x^{-2/5} \right)^9 \) is given by:

\[ T_{r+1} = \binom{9}{r} \left( x^{2/3} \right)^{9-r} \left( \frac{x^{-2/5}}{2} \right)^r \]

Simplify the expression:

\[ T_{r+1} = \binom{9}{r} \left( \frac{1}{2} \right)^r x^{\left( \frac{6 - 2r}{3} - \frac{2r}{5} \right)} \]

Step 2. For \( x^{2/3} \):
Set the power of \( x \) equal to \( 2/3 \):

\[ \frac{6 - 2r}{3} - \frac{2r}{5} = \frac{2}{3} \]

Solving this equation gives \( r = 5 \).
Substituting \( r = 5 \) into the coefficient formula:

\[ \text{Coefficient of } x^{2/3} = \binom{9}{5} \left( \frac{1}{2} \right)^5 \]

Step 3. For \( x^{-2/5} \):
Set the power of \( x \) equal to \( -2/5 \):

\[ \frac{6 - 2r}{3} - \frac{2r}{5} = -\frac{2}{5} \]

Solving this equation gives \( r = 6 \).
Substituting \( r = 6 \) into the coefficient formula:

\[ \text{Coefficient of } x^{-2/5} = \binom{9}{6} \left( \frac{1}{2} \right)^6 \]

Step 4. Sum of the Coefficients:
Add the two coefficients:

\[ \text{Sum} = \binom{9}{5} \left( \frac{1}{2} \right)^5 + \binom{9}{6} \left( \frac{1}{2} \right)^6 \]

Simplify:\[ \text{Sum} = \frac{21}{4}\]